Rolle's theorem is a special case of the mean value theorem. It is discussed here through examples and questions.
Rolle's TheoremRolle's theorem is the result of the mean value theorem where under the conditions:f(x) be a continuous functions on the interval [a, b] and differentiable on the open interval (a, b) , there exists at least one value c of x such that f '(c) = [ f(b) - f(a) ] /(b - a). Now if the condition f(a) = f(b) is satisfied, then the above simplifies to : f '(c) = 0. In other words, under the above three conditions we can always find a tangent to the curve of f that is horizontal (slope = f'(c) = 0). Rolle's Theorem
If f(x) is
1) a continuous function on the interval [a, b] 2) differentiable on the open interval (a, b) 3) and f(a) = f(b), then there is at least one value c of x in the interval (a, b) such that f '(c) = 0 Example 1The graph of f(x) = - x^{2} + 6x - 6 for 1 ≤ x ≤ 5 is shown below. f(1) = f(5) = - 1 and f is continuous on [1 , 5] and differentiable on (1 , 5) hence, according to Rolle's theorem, there exists at least one value of x = c such that f '(c) = 0.f '(x) = - 2 x + 6 f '(c) = - 2 c + 6 = 0 Solve the above equation to obtain c = 3 Therefore at x = 3 there is a tangent to the graph of f that has a slope equal to zero (horizontal line) as shown in figure 1 below. Example 2The graph of f(x) = sin(x) + 2 for 0 ≤ x ≤ 2π is shown below. f(0) = f(2π) = 2 and f is continuous on [0 , 2π] and differentiable on (0 , 2π) hence, according to Rolle's theorem, there exists at least one value (there may be more than one!) of x = c such that f '(c) = 0.f '(x) = cos(x) f '(c) = cos(c) = 0 The above equation has two solutions on the interval [0 , 2π] c_{1} = π/2 and c_{2} = 3π/2. Therefore both at x = π/2 and x = 3 π/2 there are tangents to the graph that have a slope equal to zero (horizontal line) as shown in figure 2 below. Example 3Function f in figure 3 does not satisfy Rolle's theorem: although it is continuous and f(-1) = f(3), the function is not differentiable at x = 1 and therefore f '(c) = 0 with c in the interval (-1 , 3) is not guaranteed. In fact it is easy to see that there is no horizontal tangent to the graph of f on the interval (-1 , 3).Question 1Which of the functions given below satisfy all three conditions of Rolle's theorem?a) f(x) = cos(x) , for x in [0 , 2π] b) g(x) = |x - 2| , for x in [0 , 4] c) h(x) = 1 / x^{2} , for x in [-1 , 1] d) k(x) = |sin(x)| , for x in [0 , 2π] solution to question 1a)f(0) = 1 and f(2π) = 1 therefore f(0) = f(2π) f is continuous on [0 , 2π] Function f is differentiable in (0 , 2π) Function f satisfies all conditions of Rolle's theorem b) function g has a V-shaped graph with vertex at x = 2 and is therefore not differentiable at x = 2. Function g does not satisfy all conditions of Rolle's theorem c) Function h is undefined at x = 0. Function h does not satisfy all conditions of Rolle's theorem. d) k(x) = |sin(x)| , for x in [0 , 2π] The graph of function k is shown below and it shows that function k is not differentiable at x = π. Function k does not satisfy all conditions of Rolle's theorem. Question 2Check that function f(x) = x^{ 2} - 4 x + 3 on the interval [1 , 3] satisfies all conditions of Rolle's theorem and then find all values of x = c such that f '(c) = 0.solution to question 2f is a polynomial function and is therefore continuous on the interval [1 , 3] and differentiable on the interval (1 , 3). Also f(1) = f(3) = 0 and therefore function f satisfies all three conditions of Rolle's theorem and there is at least one value of x = c such that f '(c) = 0.f '(x) = 2 x - 4 f '(c) = 2 c - 4 = 0 Solve for c to obtain c = 2. The graph below shows f , f ' and the tangent at x = c = 2 is horizontal; and f '(2) = 0 Question 3Check that function g(x) = cos(x) on the interval [- π/2 , 3π/2] satisfies all conditions of Rolle's theorem and then find all values x = c such that g '(c) = 0.solution to question 3Function g is a cosine function and is therefore continuous on the interval [- π/2 , 3π/2] and differentiable on the interval(-π/2 , 3π/2). Also g(- π/2) = g(3π/2) = 0 and therefore function g satisfies all three conditions of Rolle's theorem and there is at least one value of x = c such that f '(c) = 0. g '(x) = - sin(x) g '(c) = - sin(c) = 0 Solve for c to obtain c = nπ , n = 0,± 1 , ± 2 , ... Solutions on the interval [- π/2 , 3π/2] are c_{1} = 0 and c_{2} = π The graph below shows g , g ' and the tangents at x = c_{1} = 0 and x = c_{2} = π are horizontal; and g '(0) = 0 and g '(π) = 0 |